An object of m kg with speed of v m-s strikes a wall at an angle - and rebounds at the same speed and same angle- The magnitude of the change in momentum of the object will be

The correct option is A 2mvcosθ P⃗_1=mvsinθî mvcosθĵ and P⃗_2=mvsinθî+mvcosθĵ So change in momentum Δ P=P⃗_2 P⃗_1=2mvcosθĵ, | ΔP⃗| =2mvcosθ ...

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Standard XII

Physics

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An object of ...

Question

An object of $m k g$ with speed of $v m / s$ strikes a wall at an angle $θ$ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

2 m v cos θ

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2 m v sin θ

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2 m v

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Solution

The correct option is A $2 m v cos θ$
${\to P}_{1} = m v sin θ^i - m v cos θ^j$
and ${\to P}_{2} = m v sin θ^i + m v cos θ^j$
So change in momentum
$- - \to Δ P = {\to P}_{2} - {\to P}_{1} = 2 m v cos θ^j, | Δ \to P | = 2 m v cos θ$

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An object of m kg with speed of v m/s strikes a wall at an angle θ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

The correct option is A 2mvcosθ→P1=mvsinθ^i−mvcosθ^jand →P2=mvsinθ^i+mvcosθ^jSo change in momentum−−→ΔP=→P2−→P1=2mvcosθ^j,|Δ→P|=2mvcosθ

An object of $m k g$ with speed of $v m / s$ strikes a wall at an angle $θ$ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

The correct option is A $2 m v cos θ$
${\to P}_{1} = m v sin θ^i - m v cos θ^j$
and ${\to P}_{2} = m v sin θ^i + m v cos θ^j$
So change in momentum
$- - \to Δ P = {\to P}_{2} - {\to P}_{1} = 2 m v cos θ^j, | Δ \to P | = 2 m v cos θ$